$\begingroup$Yes, the two solutions are equivalent (with different values of $c_1$ and $c_2$). If you have the first solution, you can always re-write it in the form of the second solution, and vice versa. Yes, your last expression is the general solution.$\endgroup$
– John BarberMay 11 at 22:08

1

$\begingroup$It's simply a changing of the basis. Instead of $(e^{2x},e^{-2x})$ you take $(\cosh(2x),\sinh(2x))$. Of course coordinates, $c_1,c_2$, will change.$\endgroup$
– WangMay 11 at 22:11